If y^2 = a x^2 + bx + c, then y^3 d^2 y d x^2 is - Vidyadip

MCQ

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

✓
A constant
B
A function of $ x$ only
C
A function of $ y$ only
D
A function of $ x$ and $y$

✓

Answer

Correct option: A.

A constant

a
(a) ${y^2} = a{x^2} + bx + c \Rightarrow 2y\frac{{dy}}{{dx}} = 2ax + b$

==> $2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = 2a $

$\Rightarrow y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{dy}}{{dx}}} \right)^2}$

==>$y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{2ax + b}}{{2y}}} \right)^2}$

==> $y\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4a{y^2} - {{(2ax + b)}^2}}}{{4{y^2}}}$

==> $4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4a(a{x^2} + bx + c) - (4{a^2}{x^2} + 4abx + {b^2})$

==>$4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4ac - {b^2} \Rightarrow {y^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4ac - {b^2}}}{4} = $a constant.

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