Question

The equation of tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, having positive intercept on $y-axis$ and whose slope is ${}^{\prime }{1}^{\prime }$, is

• A. $y=x+4$
• B. $y=x+3$
• C. $y=x+5$
• D. $y=x+1$

Answer 1

Answer: (C)

$y=x+5$

We have,

$\frac{x^2}{16}+\frac{y^2}{9}=1$

$\frac{2x}{16}+\frac{2y}{9}.\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{2x}{16}.\frac{9}{2y}$

$=\frac{-9x}{16y}$

${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{-9x}{16y_1}$

As the given slope is 1.

$\frac{-9x_1}{16y_1}=1$

$-9x_1=16y_1...............\left(1\right)$

$9x_1+16y_1=0$

$x_1=\frac{-16}{9}{y}_1$

$\frac{{\left(\frac{-16}{9}{y}_1\right)}^2}{16}+\frac{y_1^2}{9}=1$

$\frac{256y_1^2}{1296}+\frac{y_1^2}{9}=1$

$y_1^2=\frac{11664}{3600}$

$y_1=\frac{108}{60}$

$y_1=\frac{9}{5}$

$\therefore x_1=\frac{-16}{5}$

Equation of tangent through $(x_1,y_1)$ with slope 1 is

$y-\frac{9}{5}=x+\frac{16}{5}$

$5y-9=5x+16$

$5y=5x+25$

$y=x+5$

Hence, this is the answer.