If the points of intersection of two distinct conics x²+y²=4b$x^2+y^2=4b$ and (x²)/(16)+(y²)/(b²)=1$\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve y²=3x²$y^2=3x^2$ then 3√(3)$3\sqrt{3}$ times the area of the rectangle formed by the intersection points is

Related Formula

Core Logic

For the given ellipse (x²)/(36) + (y²)/(16) = 1$\frac{x^2}{36} + \frac{y^2}{16} = 1$: a² = 36 ⇒ a = 6$a^2 = 36 \Rightarrow a = 6$ b² = 16$b^2 = 16$

Foci of the ellipse are at (± ae₁, 0) = (± 6 · √(5)3, 0) = (± 2√(5), 0)$(\pm ae_1, 0) = \left(\pm 6 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\pm 2\sqrt{5}, 0)$.

Step 1: Establish Hyperbola Parameters

Let the hyperbola be (x²)/(p²) - (y²)/(q²) = 1$\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$. Its foci coincide with the ellipse, so the foci of hyperbola are also (± 2√(5), 0)$(\pm 2\sqrt{5}, 0)$. Let e$e$ be the eccentricity of the hyperbola. We are given e = 5$e = 5$. Focus of hyperbola is pe = 2√(5)$pe = 2\sqrt{5}$.

Step 2: Find the Conjugate Axis (q)

For the hyperbola:

Step 3: Calculate Latus Rectum

Length of Latus Rectum = (2q²)/(p)$\frac{2q^2}{p}$

Pattern Recognition

Co-focal conics share the exact mathematical value of their focal length ae$ae$ (or pe$pe$). Instantly extract c = ae$c = ae$ from the first shape and map it directly to c = pe$c = pe$ for the second.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

For external tangents from point R$R$ forming angle 2θ$2\theta$ at the center, the distance d$d$ from center to intersection point obeys θ = (r)/(d)$\cos \theta = \frac{r}{d}$.

Core Logic

Given circle: x² + y² - 4x - 6y - 3 = 0$x^2 + y^2 - 4x - 6y - 3 = 0$ Center O = (2, 3)$O = (2, 3)$ Radius r = √((-2)² + (-3)² - (-3)) = √(4 + 9 + 3) = √(16) = 4$r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4$.

The tangents PQ and MN intersect at some point R(h, k)$R(h, k)$. The radius vectors OA$OA$ and OB$OB$ subtend angle ∠ AOB = (π)/(3) = 60°$\angle AOB = \frac{\pi}{3} = 60^{\circ}$ at the center. The line joining O$O$ and R$R$ bisects the angle ∠ AOB$\angle AOB$. Thus, ∠ AOR = 30°$\angle AOR = 30^{\circ}$.

Step 1: Apply Trigonometric Relations

In the right-angled triangle Δ AOR$\Delta AOR$, OA$OA$ is the radius (r = 4$r = 4$) and OR$OR$ is the hypotenuse.

Step 2: Construct the Locus Equation

The distance squared between O(2,3)$O(2,3)$ and R(h,k)$R(h,k)$ is OR²$OR^2$:

Step 3: Generalize the Equation

Replace (h, k)$(h, k)$ with (x, y)$(x, y)$ for the general locus:

Pattern Recognition

The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by 1/ (α/2)$1/\sin(\alpha/2)$ or 1/ (θ)$1/\cos(\theta)$ depending on whether the angle is measured at intersection or center.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

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Core Logic

Given parabola x² = 4y$x^2 = 4y$, its vertex O = (0, 0)$O = (0, 0)$. A general point Q$Q$ on x² = 4y$x^2 = 4y$ is (2t, t²)$(2t, t^2)$. Let P(h, k)$P(h, k)$ divide OQ$OQ$ in ratio 2:3$2:3$. By section formula:

Step 1: Finding the Locus C

From h = (4t)/(5)$h = \frac{4t}{5}$, we get t = (5h)/(4)$t = \frac{5h}{4}$. Substitute into k$k$:

So the conic C is the parabola 5x² = 8y$5x^2 = 8y$.

Step 2: Chord bisected at a point

We need the equation of the chord of C: 5x² - 8y = 0$C: 5x^2 - 8y = 0$ bisected at (x₁, y₁) = (1, 2)$(x_1, y_1) = (1, 2)$. Use T = S₁$T = S_1$. T = 5xx₁ - 4(y + y₁) = 5x(1) - 4(y + 2) = 5x - 4y - 8$T = 5xx_1 - 4(y + y_1) = 5x(1) - 4(y + 2) = 5x - 4y - 8$ S₁ = 5(1)² - 8(2) = 5 - 16 = -11$S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$

Equating T$T$ and S₁$S_1$:

5x - 4y + 3 = 0$5x - 4y + 3 = 0$

Pattern Recognition

Internal division locus of a vertex chord on standard conic identically scales the conic. Once the child-conic is found, standard mid-point chord protocol (T=S₁$T=S_1$) strictly applies algebraically.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Straight Lines

Related Formula

Core Logic

Step 1: Determine alpha

Rewrite the ellipse: (x²)/(1/3) + (y²)/(1/4) = 1 a² = (1)/(3), b² = (1)/(4)$\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1 \implies a^2 = \frac{1}{3}, b^2 = \frac{1}{4}$. The line is y = -(α)/(4)x + √(7)4$y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$. Using c² = a²m² + b²$c^2 = a^2m^2 + b^2$:

Since P$P$ is in the first quadrant, coordinates x, y$x, y$ are positive, so we use the tangent 3x + 4y - √(7) = 0$3x + 4y - \sqrt{7} = 0$.

Step 2: Find Point of Contact P

The tangent at P(x₁, y₁)$P(x_1, y_1)$ is 3xx₁ + 4yy₁ = 1$3xx_1 + 4yy_1 = 1$. Comparing this with 3x + 4y = √(7)$3x + 4y = \sqrt{7}$ (divided by √(7)$\sqrt{7}$ to match constant 1$1$): 3x√(7) + 4y√(7) = 1$\frac{3x}{\sqrt{7}} + \frac{4y}{\sqrt{7}} = 1$. Comparing coefficients:

So P = ( 1√(7), 1√(7))$P = \left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$.

Step 3: Calculate Focal Distance

Find eccentricity:

The focal distances are a - ex$a - ex$ and a + ex$a + ex$. Since a² = 1/3 a = 1/√(3)$a^2 = 1/3 \implies a = 1/\sqrt{3}$.

Matching with the options, the focal distance is 1√(3) + 12√(7)$\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.

Pattern Recognition

For tangency lx+my+n=0$lx+my+n=0$ to x²/a² + y²/b² = 1$x^2/a^2 + y^2/b^2 = 1$, use a² l² + b² m² = n²$a^2 l^2 + b^2 m^2 = n^2$. Points of contact can be quickly evaluated by comparing T=0$T=0$ to the normalized tangent equation.

Chapter Mix

Class 11 Maths: Conic Sections

Related Formula

Core Logic

Step 1: Locate A via Perpendicularity

Slope of OP$OP$ is mOP = (6t - 0)/(3t² - 0) = (2)/(t)$m_{OP} = \frac{6t - 0}{3t^2 - 0} = \frac{2}{t}$. Since ∠ OPA = 90^°$\angle OPA = 90^\circ$, slope of AP$AP$ is mAP = -(t)/(2)$m_{AP} = -\frac{t}{2}$. Equation of AP$AP$:

To find A$A$ on the x-axis, put y = 0$y = 0$:

So, A$A$ is (12 + 3t², 0)$(12 + 3t^2, 0)$.

Step 2: Locus of the Centroid

Let the centroid of OPA$\triangle OPA$ be G(h, k)$G(h, k)$.

From k = 2t t = (k)/(2)$k = 2t \implies t = \frac{k}{2}$. Substitute t$t$ into the equation for h$h$:

Replacing (h, k)$(h, k)$ with (x, y)$(x, y)$, the locus is y² = 2x - 8 y² - 2x + 8 = 0$y^2 = 2x - 8 \implies y^2 - 2x + 8 = 0$.

Pattern Recognition

For any right-angled configuration involving the origin and axis on a parabola, parametric geometry simplifies equations significantly. Find coordinates O, P, A$O, P, A$, set up the centroid algebraic relations, and eliminate parameter t$t$.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines