Let a tangent to Curve 9 x^2+16 y^2=144 intersect coordinate axes at points A and B . Then, minimum

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

hard

Let a tangent to the Curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $A B$ is $.........$

$5$

$6$

$7$

$8$

(JEE MAIN-2023)

Solution

Equation of tangent at point $P (4 \cos \theta, 3 \sin \theta)$ is $\frac{x \cos \theta}{4}+\frac{y \sin \theta}{3}=1$ So A is $(4 \sec \theta, 0)$ and point $B$ is $(0,3 \operatorname{cosec} \theta)$

Length $A B =\sqrt{16 \sec ^2 \theta+9 \operatorname{cosec}^2 \theta}$ $=\sqrt{25+16 \tan ^2 \theta+9 \cot ^2 \theta} \geq 7$

Standard 11

Mathematics

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is

medium

The line $x =8$ is the directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0,4 \sqrt{3})$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to $……..$

hard

(JEE MAIN-2023)

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :

normal

An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is

easy

The co-ordinates of the foci of the ellipse $3{x^2} + 4{y^2} – 12x – 8y + 4 = 0$ are