The equations of the tangents to the ellipse $9x^2 + 16y^2 = 144$ which pass through the point $(2, 3)$ are

  • A
    $y = 3, x + y = 5$
  • B
    $y = -3, x - y = 5$
  • C
    $y = 4, x + y = 3$
  • D
    $y = -4, x - y = 3$

Explore More

Similar Questions

An ellipse has its major axis along the $y$-axis and its minor axis along the $x$-axis. If the length of its latus rectum is $\frac{2}{3}$ times the length of its minor axis,then the eccentricity of the ellipse is:

The lengths of the major and minor axes of an ellipse are $10$ and $8$ respectively,and its major axis lies along the $y$-axis. The equation of the ellipse,with its center at the origin,is:

Let $E_1 = \frac{x^2}{9} + \frac{y^2}{4} = 1$ and $E_2 = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be two ellipses and $R$ be a rectangle with sides parallel to the coordinate axes. Let $E_1$ be the inscribed ellipse in $R$ and $E_2$ be the circumscribed ellipse on $R$. If $E_2$ passes through $(0, 4)$,then:

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

Let the line $y-x=1$ intersect the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at the points $A$ and $B$. Then the angle subtended by the line segment $AB$ at the center of the ellipse is:

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