Let the normal at a point P on the curve y^{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the point $P$ be $(\alpha, \beta)$. Since $P$ lies on the curve,we have $\beta^{2}-3\alpha^{2}+\beta+10=0 \dots(i)$.Differentiating the equati

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let the point $P$ be $(\alpha, \beta)$. Since $P$ lies on the curve,we have $\beta^{2}-3\alpha^{2}+\beta+10=0 \dots(i)$.Differentiating the equation of the curve with respect to $x$,we get $2yy' - 6x + y' = 0$,which implies $y'(2y+1) = 6x$,so $y' = \frac{6x}{2y+1}$.The slope of the tangent $m$ at $P(\alpha, \beta)$ is $m = \frac{6\alpha}{2\beta+1} \dots(ii)$.The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{2\beta+1}{6\alpha}$.The normal passes through $(0, \frac{3}{2})$ and $(\alpha, \beta)$,so its slope is $\frac{\beta - 3/2}{\alpha - 0} = \frac{2\beta-3}{2\alpha}$.Equating the slopes: $\frac{2\beta-3}{2\alpha} = -\frac{2\beta+1}{6\alpha}$.Assuming $\alpha 
eq 0$,we have $3(2\beta-3) = -(2\beta+1) \Rightarrow 6\beta - 9 = -2\beta - 1 \Rightarrow 8\beta = 8 \Rightarrow \beta = 1$.Substituting $\beta = 1$ into $(i)$: $1^{2} - 3\alpha^{2} + 1 + 10 = 0 \Rightarrow 3\alpha^{2} = 12 \Rightarrow \alpha^{2} = 4$.From $(ii)$,$|m| = |\frac{6\alpha}{2(1)+1}| = |\frac{6\alpha}{3}| = |2\alpha|$.Since $\alpha^{2} = 4$,$|\alpha| = 2$,thus $|m| = 2 	imes 2 = 4$.

Let the normal at a point $P$ on the curve $y^{2}-3x^{2}+y+10=0$ intersect the $y$-axis at $(0, \frac{3}{2})$. If $m$ is the slope of the tangent at $P$ to the curve,then $|m|$ is equal to

Similar Questions

If $ST$ and $SN$ are the lengths of the subtangent and the subnormal at the point $\theta = \frac{\pi}{2}$ on the curve $x = a(\theta + \sin \theta ), y = a(1 - \cos \theta ), a \neq 0$,then:

If $\theta$ is the acute angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$,then $\tan \theta =$

The sum of the intercepts on the axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is equal to:

If at any point $(x_1, y_1)$ on the curve $y=f(x)$ the lengths of the subtangent and subnormal are equal,then the length of the tangent drawn to that curve at that point is

If $\frac{k}{\alpha^3}$ is the length of the subnormal at any point $P(\alpha, y)$ on the curve $x^2-a^2=\frac{x^2 y^2}{a^2}$,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module