Let r$r$ be the radius of the circle, which touches x$x$ -axis at point (a, 0)$(a, 0)$ , a < 0$a < 0$ and the parabola y^2 = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 30 to 30+4 marks
Solution & Explanation
### Related Formula
textTangent line at point (x_1, y_1) implies yy_1 = 2a(x+x_1)$$\text{Tangent line at point } (x_1, y_1) \implies yy_1 = 2a(x+x_1)$$
### Core Logic
Establish the tangent vector expression at the parabola intersection mark. Since this path line functions as a shared contact tangent boundaries sheet for the circular arc, impose radius equations.
### Step 1: Derive Shared Parabola Tangent Line
Tangent line profile for y^2 = 9x$y^2 = 9x$ at coordinate indicator (4,6)$(4,6)$:
6y = 9 cdot left( fracx+42 right) implies 3x - 4y + 12 = 0$$6y = 9 \cdot \left( \frac{x+4}{2} \right) \implies 3x - 4y + 12 = 0$$
### Step 2: Build Geometric Metric Connections
Circle touches axis at (a,0)$(a,0)$, mapping coordinates center directly to C(a,r)$C(a,r)$. Perpendicular boundary constraint steps require:
frac3a - 4r + 125 = pm r implies 3a + 12 = 4r pm 5r$$\frac{3a - 4r + 12}{5} = \pm r \implies 3a + 12 = 4r \pm 5r$$
### Step 3: Solve for Radius Matrix Bounds
Enforce circle equation intersection constraint profile (x-a)^2 + (y-r)^2 = r^2$(x-a)^2 + (y-r)^2 = r^2$ at point (4,6)$(4,6)$:
a^2 - 8a - 12r + 52 = 0$$a^2 - 8a - 12r + 52 = 0$$
Evaluating the target systems from structural logic tracks rejects positive value parameters, providing:
a = -14, quad r = 30$$a = -14, \quad r = 30$$
{{SOL_IMG_75}}
### Pattern Recognition
Shared tangent elements connect independent conic fields. Locating circular center boundaries using axial coordinate tracking simplifies secondary equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Circles
Tangent to Parabola and Circle Properties diagram for Q75 - JEE Main 2025 Evening
The centre of a circle C$C$ is at the centre of the ellipse E: fracx^2a^2 + fracy^2b^2 = 1, a > b$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$. Let C$C$ pass through the foci F_1$F_1$ and F_2$F_2$ of E such that the circle $E such that the circle $C and the ellipse $ and the ellipse $E intersect at four points. Let $ intersect at four points. Let $P be one of these four points. If the area of the triangle $ be one of these four points. If the area of the triangle $PF_1F_2 is 30 and the length of the major axis of $ is 30 and the length of the major axis of $E is 17, then the distance between the foci of $ is 17, then the distance between the foci of $E$ is :
A.26$26$
B.13$13$
C.12$12$
D.frac132$\frac{13}{2}$
Solution
### Core Logic
The circle C$C$ has its center at the origin and passes through the foci F_1(-ae, 0)$F_1(-ae, 0)$ and F_2(ae, 0)$F_2(ae, 0)$. This means F_1F_2$F_1F_2$ is the diameter of the circle.
Any point P$P$ lying on this circle satisfies the property that the angle subtended by the diameter is a right angle:
angle F_1PF_2 = 90^circ$$\angle F_1PF_2 = 90^\circ$$Ellipse properties diagram for Q67 - JEE Main 2025 Evening
### Step 1: Using the Area and Ellipse Definition
Since triangle PF_1F_2$\triangle PF_1F_2$ is a right-angled triangle at P$P$:
textArea = frac12 cdot PF_1 cdot PF_2 = 30 implies PF_1 cdot PF_2 = 60$$\text{Area} = \frac{1}{2} \cdot PF_1 \cdot PF_2 = 30 \implies PF_1 \cdot PF_2 = 60$$
By the definition of an ellipse, the sum of the focal distances to any point on the curve is equal to the length of the major axis (2a = 17$2a = 17$):
PF_1 + PF_2 = 17$$PF_1 + PF_2 = 17$$
### Step 2: Calculating Distance between Foci
Applying Pythagoras' theorem in right-angled triangle PF_1F_2$\triangle PF_1F_2$:
F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 cdot PF_2)$$F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 \cdot PF_2)$$
Substitute the known values from our equations block:
F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$$F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169$$F_1F_2 = sqrt169 = 13$$F_1F_2 = \sqrt{169} = 13$$
Therefore, the distance between the foci is 13$13$.
### Pattern Recognition
Whenever a circle is circumscribed around the foci of an ellipse, remember Thales' theorem: any intersection point with the ellipse forms a right triangle with the focal diameter, linking focal properties directly to Pythagoras.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q67jee_main_2025_04_april_morningEllipse - Foci and Latus Rectum
The length of the latus-rectum of the ellipse, whose foci are (2, 5)$(2, 5)$ and (2, -3)$(2, -3)$ and eccentricity is frac45$\frac{4}{5}$, is
A.frac65$\frac{6}{5}$
B.frac503$\frac{50}{3}$
C.frac103$\frac{10}{3}$
D.frac185$\frac{18}{5}$
Solution
### Related Formula
Distance between foci = 2be$= 2be$ (for vertical major axis).
Length of Latus Rectum (textL.R.) = frac2a^2b$(\text{L.R.}) = \frac{2a^2}{b}$.
### Core Logic
Foci are F_1(2,5)$F_1(2,5)$ and F_2(2,-3)$F_2(2,-3)$. The x$x$-coordinates are identical, indicating a vertical ellipse.
Distance between foci:
2be = 5 - (-3) = 8 implies be = 4$$2be = 5 - (-3) = 8 \implies be = 4$$
Given eccentricity e = frac45$e = \frac{4}{5}$:
bleft(frac45right) = 4 implies b = 5$$b\left(\frac{4}{5}\right) = 4 \implies b = 5$$Ellipse - Foci and Latus Rectum diagram for Q67 - JEE Main 2025 Morning
### Step 1: Calculate Minor Axis length
Use eccentricity relation:
a^2 = b^2(1 - e^2) implies a^2 = 25left(1 - frac1625right) = 25 times frac925 = 9 implies a = 3$$a^2 = b^2(1 - e^2) \implies a^2 = 25\left(1 - \frac{16}{25}\right) = 25 \times \frac{9}{25} = 9 \implies a = 3$$
### Step 2: Evaluate Latus Rectum
textL.R. = frac2a^2b = frac2 times 95 = frac185$$\text{L.R.} = \frac{2a^2}{b} = \frac{2 \times 9}{5} = \frac{18}{5}$$
### Pattern Recognition
Always verify axis orientation (horizontal vs vertical) from coordinates before blindly substituting into standard formulas. Symmetrical components match axis lengths parameters directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q74jee_main_2025_04_april_morningTangents to Conics
Let C be the circle x^2 + (y - 1)^2 = 2$x^2 + (y - 1)^2 = 2$, E_1$E_1$ and E_2$E_2$ be two ellipses whose centres lie at the origin and major axes lie on the x-axis and y-axis respectively. Let the straight line x + y = 3$x + y = 3$ touch the curves C, E_1$E_1$ and E_2$E_2$ at P(x_1, y_1)$P(x_1, y_1)$, Q(x_2, y_2)$Q(x_2, y_2)$ and R(x_3, y_3)$R(x_3, y_3)$ respectively. Given that P is the mid-point of the line segment QR and PQ = frac2sqrt23$PQ = \frac{2\sqrt{2}}{3}$, the value of 9(x_1y_1 + x_2y_2 + x_3y_3)$9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to
Numerical Answer.Answer: 46 to 46
Solution
### Related Formula
Parametric equation of a straight line:
x = x_1 + rcostheta, quad y = y_1 + rsintheta$$x = x_1 + r\cos\theta, \quad y = y_1 + r\sin\theta$$
### Core Logic
Step 1: Find point P(x_1,y_1)$P(x_1,y_1)$ on circle C$C$.
Equation of tangent at P$P$ on x^2 + y^2 - 2y - 1 = 0$x^2 + y^2 - 2y - 1 = 0$ is xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$.
Comparing with line x + y = 3 implies fracx_11 = fracy_1 - 11 = fracy_1 + 13$x + y = 3 \implies \frac{x_1}{1} = \frac{y_1 - 1}{1} = \frac{y_1 + 1}{3}$.
Solving gives x_1 = 1, y_1 = 2$x_1 = 1, y_1 = 2$. Thus, P = (1, 2)$P = (1, 2)$.
### Step 1: Use Line Parametrics for Q and R
Line x + y = 3$x + y = 3$ makes an angle theta = 135^circ$\theta = 135^{\circ}$ with the positive x-axis.
Using parametric distances from P(1,2)$P(1,2)$ with r = PQ = frac2sqrt23$r = PQ = \frac{2\sqrt{2}}{3}$:
x = 1 pm rcos(135^circ) = 1 mp fracrsqrt2$$x = 1 \pm r\cos(135^{\circ}) = 1 \mp \frac{r}{\sqrt{2}}$$y = 2 pm rsin(135^circ) = 2 pm fracrsqrt2$$y = 2 \pm r\sin(135^{\circ}) = 2 \pm \frac{r}{\sqrt{2}}$$
Substitute r = frac2sqrt23$r = \frac{2\sqrt{2}}{3}$:
For Q$Q$: x_2 = 1 + frac23 = frac53, y_2 = 2 - frac23 = frac43$x_2 = 1 + \frac{2}{3} = \frac{5}{3}, y_2 = 2 - \frac{2}{3} = \frac{4}{3}$.
For R$R$: x_3 = 1 - frac23 = frac13, y_3 = 2 + frac23 = frac83$x_3 = 1 - \frac{2}{3} = \frac{1}{3}, y_3 = 2 + \frac{2}{3} = \frac{8}{3}$.
### Step 2: Evaluate Final Expression
Calculate the products:
x_1y_1 = 1 times 2 = 2$x_1y_1 = 1 \times 2 = 2$x_2y_2 = frac53 times frac43 = frac209$x_2y_2 = \frac{5}{3} \times \frac{4}{3} = \frac{20}{9}$x_3y_3 = frac13 times frac83 = frac89$x_3y_3 = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9}$9(x_1y_1 + x_2y_2 + x_3y_3) = 9left(2 + frac209 + frac89right) = 18 + 20 + 8 = 46$$9(x_1y_1 + x_2y_2 + x_3y_3) = 9\left(2 + \frac{20}{9} + \frac{8}{9}\right) = 18 + 20 + 8 = 46$$
### Pattern Recognition
Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations a^2, b^2$a^2, b^2$ completely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Circles
Class 11 Mathematics: Conic Sections
Q62jee_main_2025_07_april_eveningProperties of Ellipse
Let the length of a latus rectum of an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be 10. If its eccentricity is the minimum value of the function f(t) = t^2 + t + frac1112, t in mathbfR$f(t) = t^2 + t + \frac{11}{12}, t \in \mathbf{R}$, then a^2 + b^2$a^2 + b^2$ is equal to:
A.125$125$
B.126$126$
C.120$120$
D.115$115$
Solution
### Related Formula
Length of latus rectum of an ellipse and its eccentricity relation are:
textLR = frac2b^2a$$\text{LR} = \frac{2b^2}{a}$$e^2 = 1 - fracb^2a^2$$e^2 = 1 - \frac{b^2}{a^2}$$
### Core Logic
Given length of textLR = 10 implies frac2b^2a = 10 implies b^2 = 5a quad dots text(i)$\text{LR} = 10 \implies \frac{2b^2}{a} = 10 \implies b^2 = 5a \quad \dots \text{(i)}$
Now, let's find the minimum value of f(t) = t^2 + t + frac1112$f(t) = t^2 + t + \frac{11}{12}$.
Differentiating: f'(t) = 2t + 1 = 0 implies t = -frac12$f'(t) = 2t + 1 = 0 \implies t = -\frac{1}{2}$.
textMinimum value e = fleft(-frac12
ight) = left(-frac12
ight)^2 + left(-frac12
ight) + frac1112 = frac14 - frac12 + frac1112 = frac3 - 6 + 1112 = frac812 = frac23$$\text{Minimum value } e = f\left(-\frac{1}{2}
ight) = \left(-\frac{1}{2}
ight)^2 + \left(-\frac{1}{2}
ight) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}$$
### Step 1: Solve for a and b
Using eccentricity formula:
e^2 = frac49 = 1 - fracb^2a^2 implies fracb^2a^2 = frac59 implies b^2 = frac5a^29 quad dots text(ii)$$e^2 = \frac{4}{9} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{5}{9} \implies b^2 = \frac{5a^2}{9} \quad \dots \text{(ii)}$$
Equating (i) and (ii):
5a = frac5a^29 implies a = 9$$5a = \frac{5a^2}{9} \implies a = 9$$
Then from (i):
b^2 = 5(9) = 45 implies b = 3sqrt5$$b^2 = 5(9) = 45 \implies b = 3\sqrt{5}$$
Hence, a^2 = 81$a^2 = 81$.
### Step 2: Calculate a^2 + b^2
a^2 + b^2 = 81 + 45 = 126$$a^2 + b^2 = 81 + 45 = 126$$
### Pattern Recognition
A quadratic function at^2+bt+c$at^2+bt+c$ reaches its extreme value at t = -fracb2a$t = -\frac{b}{2a}$. Using this layout avoids full calculus derivation steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Quadratic Equations
Q68jee_main_2025_07_april_eveningEccentricity and Foci
Let mathbfe_1$\mathbf{e}_1$ and mathbfe_2$\mathbf{e}_2$ be the eccentricities of the ellipse fracmathrmx^2mathrmb^2 + fracmathrmy^225 = 1$\frac{\mathrm{x}^2}{\mathrm{b}^2} + \frac{\mathrm{y}^2}{25} = 1$ and the hyperbola fracmathrmx^216 - fracmathrmy^2mathrmb^2 = 1$\frac{\mathrm{x}^2}{16} - \frac{\mathrm{y}^2}{\mathrm{b}^2} = 1$, respectively. If mathrmb < 5$\mathrm{b} < 5$ and mathrme_1mathrme_2 = 1$\mathrm{e}_1\mathrm{e}_2 = 1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:
A.frac45$\frac{4}{5}$
B.frac35$\frac{3}{5}$
C.fracsqrt74$\frac{\sqrt{7}}{4}$
D.fracsqrt32$\frac{\sqrt{3}}{2}$
Solution
### Related Formula
Eccentricity for ellipse (a$a) and hyperbola are calculated as:
e_1^2 = 1 - fraca^2b^2$$e_1^2 = 1 - \frac{a^2}{b^2}$$e_2^2 = 1 + fracb^2a^2$$e_2^2 = 1 + \frac{b^2}{a^2}$$
### Core Logic
Since b < 5$b < 5$, for the ellipse fracx^2b^2 + fracy^225 = 1$\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the major axis is along the y$y$-axis:
e_1^2 = 1 - fracb^225$$e_1^2 = 1 - \frac{b^2}{25}$$
For the hyperbola fracx^216 - fracy^2b^2 = 1$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$:
e_2^2 = 1 + fracb^216$$e_2^2 = 1 + \frac{b^2}{16}$$
Given e_1^2 e_2^2 = 1$e_1^2 e_2^2 = 1$:
left(1 - fracb^225right)left(1 + fracb^216right) = 1$$\left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1$$1 + fracb^216 - fracb^225 - fracb^4400 = 1 implies frac9b^2400 = fracb^4400 implies b^2 = 9$$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \implies \frac{9b^2}{400} = \frac{b^4}{400} \implies b^2 = 9$$
### Step 1: Calculate Foci Locations
Substituting b^2 = 9$b^2 = 9$:
Ellipse foci: ae_1 = 5 cdot sqrt1 - frac925 = 5 cdot frac45 = 4$ae_1 = 5 \cdot \sqrt{1 - \frac{9}{25}} = 5 \cdot \frac{4}{5} = 4$. Foci lie along y$y$-axis: (0, pm 4)$(0, \pm 4)$.
Hyperbola foci: ae_2 = 4 cdot sqrt1 + frac916 = 4 cdot frac54 = 5$ae_2 = 4 \cdot \sqrt{1 + \frac{9}{16}} = 4 \cdot \frac{5}{4} = 5$. Foci lie along x$x$-axis: (pm 5, 0)$(\pm 5, 0)$.
### Step 2: Construct the New Ellipse
The new ellipse passes through (0, pm 4)$(0, \pm 4)$ and (pm 5, 0)$(\pm 5, 0)$. Thus, its semi-major axis is A = 5$A = 5$ along the x$x$-axis and semi-minor axis is B = 4$B = 4$ along the y$y$-axis:
E = sqrt1 - fracB^2A^2 = sqrt1 - frac1625 = frac35$$E = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$
### Pattern Recognition
When a conic passes through points directly on the axes like (pm alpha, 0)$(\pm alpha, 0)$ and (0, pm beta)$(0, \pm \beta)$, those points immediately represent the semi-axes values A$A$ and B$B$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
More Conic Sections Questions — jee_main_2025_08_april_evening
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