In the line α x + 4y = √(7)$\alpha x + 4y = \sqrt{7}$, where α in R$\alpha \in R$, touches the ellipse3x² + 4y² = 1$3x^{2} + 4y^{2} = 1$ at the point P$P$ in the first quadrant, then one of the focal distances of P$P$ is:
Condition of tangency for ellipse (x²)/(a²) + (y²)/(b²) = 1 is c² = a²m² + b²$$\text{Condition of tangency for ellipse } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ is } c^2 = a^2m^2 + b^2$$Focal distance SP = a ± ex$$\text{Focal distance } SP = a \pm ex$$Eccentricity e = √(1 - (b²)/(a²))$$\text{Eccentricity } e = \sqrt{1 - \frac{b^2}{a^2}}$$
Core Logic
Ellipse diagram for Q4 - JEE Main 2026 Evening
Identify the slope and intercepts of the line to find α$\alpha$. Use the point of contact formula to locate P(x₁, y₁)$P(x_1, y_1)$ and apply focal distance definitions.
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$c^2 = a^2m^2 + b^2$:
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.
Keywords:#touches the ellipse#JEE Main 2026 Evening Q4#Conic Sections JEE Main 2026#Ellipse JEE Main 2026
More Conic Sections Previous-Year Questions — Page 6
Q3jee_main_2026_28_january_eveningParabola and Triangles
Let A$A$ be the focus of the parabolay² = 8x$y^{2} = 8x$. Let the line y = mx + c$y = mx + c$ intersect the parabola at two distinct points B$B$ and C$C$. If the centroid of the triangle ABC$ABC$ is ((7)/(3), (4)/(3))$\left(\frac{7}{3}, \frac{4}{3}\right)$, then (BC)²$(BC)^{2}$ is equal to :
Focus of y² = 8x$y^2 = 8x$ is A(2, 0)$A(2, 0)$ since 4a = 8 ⇒ a=2$4a = 8 \Rightarrow a=2$.
Let points on parabola be B(2t₁², 4t₁)$B(2t_1^2, 4t_1)$ and C(2t₂², 4t₂)$C(2t_2^2, 4t_2)$.
The centroid of Δ ABC$\Delta ABC$ is given as ((7)/(3), (4)/(3))$\left(\frac{7}{3}, \frac{4}{3}\right)$.
Equating coordinates:
Using parametric coordinates (at², 2at)$(at^2, 2at)$ systematically reduces algebraic complexity when determining intersections or triangle properties on a parabola.
Chapter Mix
Class 11 Maths: Conic Sections
Q9jee_main_2026_28_january_eveningEllipse Parameters and Latus Rectum
An ellipse has its center at (1,-2)$(1,-2)$, one focus at (3,-2)$(3,-2)$ and one vertex at (5, - 2)$(5, - 2)$. Then the length of its latus rectum is :
From the given coordinates on the major axis (y = -2$y = -2$):
Center C(1, -2)$C(1, -2)$, Focus F₁(3, -2)$F_1(3, -2)$, Vertex A₁(5, -2)$A_1(5, -2)$.
Distance from center to vertex, CA₁ = a = 5 - 1 = 4$CA_1 = a = 5 - 1 = 4$.
Distance from center to focus, CF₁ = ae = 3 - 1 = 2$CF_1 = ae = 3 - 1 = 2$.
Ellipse dimensions mapped to coordinates
Execution
Calculate eccentricity e$e$:
ae = 2 ⇒ 4e = 2 ⇒ e = (1)/(2)$$ae = 2 \Rightarrow 4e = 2 \Rightarrow e = \frac{1}{2}$$
Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters (a$a$ and ae$ae$) directly.
Chapter Mix
Class 11 Maths: Conic Sections
Q10jee_main_2026_28_january_eveningConfocal Ellipse and Hyperbola
Let the ellipse E: x²144 + y²169 = 1$E: \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola H: x²16 - y²λ² = -1$H: \frac{x^{2}}{16} - \frac{y^{2}}{\lambda^{2}} = -1$have the same foci. If e$e$ and L$L$ respectively denote the eccentricity and the length of the latus rectum of H$H$, then the value of 24(e + L)$24(e + L)$ is:
Eccentricity of hyperbola, e = (5)/(3)$e = \frac{5}{3}$.
Length of latus rectum of hyperbola, L = (2(16))/(λ) = (32)/(3)$L = \frac{2(16)}{\lambda} = \frac{32}{3}$.
Confocal conics usually align along the same major axis. Notice the -1$-1$ on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the b>a$b>a$ ellipse.
Chapter Mix
Class 11 Maths: Conic Sections
Q12jee_main_2026_28_january_eveningParametric Form and Chord Intersections
Let the circle x² + y² = 4$x^{2} + y^{2} = 4$ intersect x-axis at the points A(a, 0), a > 0$A(a, 0), a > 0$ and B(b, 0)$B(b, 0)$. Let P(2 α, 2 α), 0 < α < (π)/(2)$P(2 \cos\alpha, 2 \sin\alpha), 0 < \alpha < \frac{\pi}{2}$ and Q(2 β, 2 β)$Q(2 \cos\beta, 2 \sin\beta)$ be two points such that (α - β) = (π)/(2)$(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of AQ$AQ$ and BP$BP$ lies on:
Intersection of circle with x-axis provides A(2,0)$A(2,0)$ and B(-2,0)$B(-2,0)$.
Let the point of intersection of AQ$AQ$ and BP$BP$ be R(h, k)$R(h, k)$.
Since R$R$ lies on BP$BP$, the slope mBR = mBP$m_{BR} = m_{BP}$:
Locus of R$R$ is x² + y² - 4y - 4 = 0$x^2 + y^2 - 4y - 4 = 0$.
Pattern Recognition
Connecting chords from extreme diameter vertices to points whose parametric angles differ by π/2$\pi/2$ reliably generates perpendicular-like slope products or standard tangent angle identities, mapping directly to a circular locus.
Chapter Mix
Class 11 Maths: Circles
Q55jee_main_2025_02_april_eveningEllipse
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :
A.4√(17)$\frac{4}{\sqrt{17}}$
B.√(3)16$\frac{\sqrt{3}}{16}$
C.3√(19)$\frac{3}{\sqrt{19}}$
D.√(5)7$\frac{\sqrt{5}}{7}$
Solution
Related Formula
Length of minor axis = 2b$$\text{Length of minor axis} = 2b$$Distance between foci = 2ae$$\text{Distance between foci} = 2ae$$Eccentricity: e = √(1 - (b²)/(a²))$$\text{Eccentricity: } e = \sqrt{1 - \frac{b^2}{a^2}}$$
Core Logic
We set up an algebraic equation relating b$b$, a$a$, and e$e$ from the given geometric condition, then substitute it into the eccentricity identity.
Step 1: Set up the geometric relation
Given that 2b = (1)/(4) (2ae)$2b = \frac{1}{4} (2ae)$:
Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a$b/a$ as a function of e$e$ allows direct solving of the eccentricity.
Chapter Mix
Class 11 Mathematics: Conic Sections
More Conic Sections Questions — jee_main_2026_21_jan_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
