The differential equation whose solution is (x -h)^2 + (y -k)^2 =… - Vidyadip

MCQ

The differential equation whose solution is $(x -h)^2 + (y -k)^2 = a^2$ is ($a$ is a constant)

A
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}\frac{{{d^2}y}}{{d{x^2}}}$
✓
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$
C
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$
D
None of these

✓

Answer

Correct option: B.

${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {a^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

b
Given, $(x-h)^{2}+(y-k)^{2}=a^{2}$

$\Rightarrow 2(x-h)+2(y-k) \frac{d y}{d x}=0$

$\Rightarrow(x-h)+(y-k) \frac{d y}{d x}=0 \quad \ldots$ (ii)

Again differentiating $(y-k)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$

Putting in Eq. (ii), we get $x-h=-(y-k) \frac{d y}{d x}$

$=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d x^{2}}}$

Putting in Eq. (i), we get $=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=a^{2}$

$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$

$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=a^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If the function $f(x)=x^3+e^{\frac{x}{2}}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime}(1)$ is

In the set $A = \{1, 2, 3, 4, 5\}$, a relation $R$ is defined by $R = \{(x, y)| x, y$ $ \in A$ and $x < y\}$. Then $R$ is

Tangents are drawn from any point on hyperbola $4x^2 -9y^2 = 36$ to the circle $x^2 + y^2 = 9$ . If locus of midpoint of chord of contact is $\left( {\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4}} \right) = \lambda {\left( {\frac{{{x^2} + {y^2}}}{9}} \right)^2}$ , then $\lambda $ is

The equation of the circle of radius $5$ and touching the coordinate axes in third quadrant is

If $\sin \alpha + \sin \beta + \sin \gamma = 0 = $$\cos \alpha + \cos \beta + \cos \gamma ,$ then the value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $ is

Let $f\left( x \right) = \frac{x}{{\sqrt {{a^2} + {x^2}} }} - \frac{{d - x}}{{\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},x \in R\,$, where $a, b$ and $d$ are non -zero real constant. Then

If $y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to :

The chance of getting a doublet with $2$ dice is

If in a parallelogram $ABDC$, the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$, then the equation of the diagonal $AD$ is

Let $a,\,\,b$ and $c$ be non-zero vectors such that $(a \times b) \times c = \frac{1}{3}|b||c|a.$ If $\theta$ is the acute angle between the vectors $b$ and $c$, then $\sin \theta $ equals