If a tangent with slope -4/3 to the ellipse f | vedclass

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

Question

(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$.The equation of a tangent with slope $m$ to t

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$.The equation of a tangent with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given $m = -4/3$,$a^2 = 18$,and $b^2 = 32$.Substituting these values: $y = -\frac{4}{3}x \pm \sqrt{18(-\frac{4}{3})^2 + 32} = -\frac{4}{3}x \pm \sqrt{18(\frac{16}{9}) + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm \sqrt{64} = -\frac{4}{3}x \pm 8$.Taking the positive intercept case,$y = -\frac{4}{3}x + 8$.To find $A$ (intersection with the major axis,which is the $y$-axis here since $b^2 > a^2$),set $x = 0$: $y = 8$,so $A = (0, 8)$.To find $B$ (intersection with the minor axis,which is the $x$-axis),set $y = 0$: $0 = -\frac{4}{3}x + 8 \implies \frac{4}{3}x = 8 \implies x = 6$,so $B = (6, 0)$.The area of $\Delta OAB = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes 8 	imes 6 = 24$ square units.

If a tangent with slope $-4/3$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major axis and minor axis at $A$ and $B$ respectively,then the area of $\Delta OAB$ is .......... square units.

Similar Questions

If a tangent of slope $2$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$,then the maximum value of $ab$ is

The length of the latus rectum of the ellipse $16x^2 + 25y^2 = 400$ is

The locus of the point $P(x, y)$ satisfying the relation $\sqrt{(x - 3)^2 + (y - 1)^2} + \sqrt{(x + 3)^2 + (y - 1)^2} = 6$ is

The number of real tangents that can be drawn to the ellipse $3x^2 + 5y^2 = 32$ passing through $(3, 5)$ is

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to $......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module