Question

#ANS-D

Tangent at $\left(\sqrt{3 , \frac{1}{2}}\right)$ is $\sqrt{3}x + 2y = 4$

Since slope of tangent at $\left(\sqrt{3 , \frac{1}{2}}\right)$ is -ve hence possible curve are (I), (II) only

equation of curve is $\frac{x^2}{4}+y^2=1, i.e., x^2+a^2{y}^2=a^2$

Hence equation of tangent i$y=mx+Â \sqrt{a^2{m}^2+1}$

And point of contact is $(\frac{-a^{2}m}{\sqrt{a^2{m}^2+1}}, \frac{1}{\sqrt{a^2{m}^2+1}})$

#ANS-A

Tangent at (8, 16) is y = x + 8
Slope of tangent is +ve hence possible curve will be y = 4ax
$⇒$ equation of tangent is my = m²x + a and point of contact is ^{$(\frac{a}{m^2}, \frac{2a}{m^{}})$}

^{# ANS-D}

For $a=\sqrt{2}$ and point (–1, 1) on the curve is x² + y² = a² equation of tangent is $y = mx + a\sqrt{ m^2 +1}$ and point of contact is $(\frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}})$