The equation of the tangent to the curve 9x^{ | vedclass

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

Question

(A) The given curve is $9x^{2} + 16y^{2} = 144$. Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$. Here,$a^{2} = 16$ and $b^{2} = 9$.

Vedclass · Accepted Answer

$x + y = 5$

Vedclass · Answer

(A) The given curve is $9x^{2} + 16y^{2} = 144$. Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$. Here,$a^{2} = 16$ and $b^{2} = 9$. Let the equation of the tangent be $x + y = k$,which can be written as $y = -x + k$. Comparing this with $y = mx + c$,we have $m = -1$ and $c = k$. The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$. Substituting the values,we get $k^{2} = 16(-1)^{2} + 9$. $k^{2} = 16 + 9 = 25$. $k = \pm 5$. Thus,the equations of the tangents are $x + y = 5$ and $x + y = -5$.

The equation of the tangent to the curve $9x^{2} + 16y^{2} = 144$ which makes equal intercepts with the coordinate axes is:

Similar Questions

If $c \in \mathbb{R}$ be such that the line $4x - y + c = 0$ touches the ellipse $x^2 + 4y^2 = 4$,then an equation having all such values of $c$ among its roots is

If the line $y = mx + c$ touches the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,then $c = $

The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3x^2 + 2y^2 = 5$ is:

For the ellipse $\frac{x^2}{64} + \frac{y^2}{28} = 1$,the eccentricity is

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module