For what value of lambda does the line y = x | vedclass

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

Question

(C) The equation of the ellipse is $9x^2 + 16y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Comparing this with $\frac{x^2}{

Vedclass · Accepted Answer

$\pm 5$

Vedclass · Answer

(C) The equation of the ellipse is $9x^2 + 16y^2 = 144$.Dividing by $144$,we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$.Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 9$.The line equation is $y = x + \lambda$,which is of the form $y = mx + c$,where $m = 1$ and $c = \lambda$.The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.Substituting the values,we get $\lambda^2 = 16(1)^2 + 9$.$\lambda^2 = 16 + 9 = 25$.Therefore,$\lambda = \pm 5$.

For what value of $\lambda$ does the line $y = x + \lambda$ touch the ellipse $9x^2 + 16y^2 = 144$?

Similar Questions

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be equal to the length of its semi-major axis. If the radius of its director circle is $\sqrt{3}$ and $e$ is its eccentricity,then the length of its latus rectum is

If the normal at one end of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through one end of the major axis,then:

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the coordinate axes is

If the normal drawn at one end of the latus rectum of the ellipse $b^2 x^2 + a^2 y^2 = a^2 b^2$ with eccentricity $e$ passes through one end of the minor axis,then:

On the ellipse $4x^{2} + 9y^{2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module