(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{6} = 1$,where $a^2 = 4$ and $b^2 = 6$. The locus of the intersection of perpendicular tangents to

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{6} = 1$,where $a^2 = 4$ and $b^2 = 6$. The locus of the intersection of perpendicular tangents to an ellipse is its director circle,given by $x^2 + y^2 = a^2 + b^2$. Here,$a^2 = 4$ and $b^2 = 6$. Therefore,the equation of the director circle is $x^2 + y^2 = 4 + 6 = 10$. Since $(\alpha, \beta)$ is the point of intersection of two perpendicular tangents,it must lie on the director circle. Thus,$\alpha^2 + \beta^2 = 10$.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Ellipse

Medium

Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

AP EAMCET 2024 All AP EAMCET

A
$10$
B
$52$
C
$26$
D
$5/12$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

10-2. Parabola, Ellipse, Hyperbola

Subtopic

Ellipse

Previous Year

AP EAMCET 2024 Paper All AP EAMCET years →

Find the locus of the midpoint of the portion of any tangent to the ellipse $x^{2} + 2y^{2} = 2$ intercepted between the axes.

Difficult

View Solution

If the eccentricity of the ellipse $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$ is $\frac{1}{\sqrt{6}}$,find the length of the latus rectum of the ellipse.

Medium

View Solution

If the distance of a point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ from the center is $2$,find its eccentric angle $\varphi$.

Medium

View Solution

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

DifficultTS EAMCET 2011

View Solution

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices,then the midpoints of the tangents intercepted between the coordinate axes lie on the curve:

DifficultJEE MAIN 2019

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

Similar Questions

Find the locus of the midpoint of the portion of any tangent to the ellipse $x^{2} + 2y^{2} = 2$ intercepted between the axes.

If the eccentricity of the ellipse $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$ is $\frac{1}{\sqrt{6}}$,find the length of the latus rectum of the ellipse.

If the distance of a point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ from the center is $2$,find its eccentric angle $\varphi$.

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices,then the midpoints of the tangents intercepted between the coordinate axes lie on the curve:

Vedclass Test Series

Exam Paper Generator

Online Exam Module