Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Write the correct alternative in the following:

Question:

Write the correct alternative in the following:

If $y^{2}=a x^{2}+b x+c$, then $y^{3} \frac{d^{2} y}{d x^{2}}$ is

A. a constant

B. a function of $x$ only

C. a function of $y$ only

D. a function of $x$ and $y$

Solution:

Given:

$y^{2}=a x^{2}+b x+c$

$\left.y=\sqrt{(} a x^{2}+b x+c\right)$

$\frac{d y}{d x}=\frac{1}{2 \sqrt{\left(a x^{2}+b x+c\right)}} \times(2 a x+b)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

$=\frac{1}{2}\left\{\frac{\left(2 a \times \sqrt{a x^{2}+b x+c}\right)-\left((2 a x+b) \times \frac{1}{\left.2 \sqrt{(} a x^{2}+b x+c\right)} \times(2 a x+b)\right)}{\left.\left(\sqrt{(} a x^{2}+b x+c\right)\right)^{2}}\right\}$

$=\frac{1}{2}\left\{\frac{\frac{4 a\left(a x^{2}+b x+c\right)-(2 a x+b)^{2}}{2 \sqrt{\left(a x^{2}+b x+c\right.}}}{\left(\sqrt{\left.\left(a x^{2}+b x+c\right)\right)^{2}}\right.}\right\}$

$=\frac{1}{2}\left\{\frac{4 a^{2} x^{2}+4 a b x+4 a c-4 a^{2} x^{2}-b^{2}-4 a b x}{\left.\left(\sqrt{(} a x^{2}+b x+c\right)\right)^{2} \times 2 \sqrt{(} a x^{2}+b x+c}\right\}$

$=\frac{1}{4}\left\{\frac{4 a c-b^{2}}{\left.\left(\sqrt{(} a x^{2}+b x+c\right)\right)^{\frac{3}{2}}}\right\}$

$\left.\mathrm{y}^{3} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{4}\left\{\frac{4 \mathrm{ac}-\mathrm{b}^{2}}{\left.\left(\sqrt{(} \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right)^{3}}\right\} \times\left(\sqrt{(} \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right)^{3}$

$=\frac{4 a c-b^{2}}{4}$

Hence, $y$ is a constant.