The normal to the curve x^{2}=4y passing thro | vedclass

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

Question

(D) The equation of the given curve is $x^{2}=4y$.Differentiating with respect to $x$,we have:$2x = 4 \cdot \frac{dy}{dx}$$\Rightarrow \frac{dy}{dx} =

Vedclass · Accepted Answer

$x+y=3$

Vedclass · Answer

(D) The equation of the given curve is $x^{2}=4y$.Differentiating with respect to $x$,we have:$2x = 4 \cdot \frac{dy}{dx}$$\Rightarrow \frac{dy}{dx} = \frac{x}{2}$.The slope of the normal to the curve at point $(h, k)$ is given by:$m = -\frac{1}{\left(\frac{dy}{dx}\right)_{(h,k)}} = -\frac{2}{h}$.The equation of the normal at point $(h, k)$ is:$y - k = -\frac{2}{h}(x - h)$.Since the normal passes through $(1, 2)$,we have:$2 - k = -\frac{2}{h}(1 - h) \Rightarrow k = 2 + \frac{2}{h}(1 - h) \dots (i)$.Since $(h, k)$ lies on the curve $x^{2} = 4y$,we have $h^{2} = 4k$,which implies $k = \frac{h^{2}}{4}$.Substituting $k$ in equation $(i)$:$\frac{h^{2}}{4} = 2 + \frac{2}{h} - 2$$\frac{h^{2}}{4} = \frac{2}{h}$$h^{3} = 8 \Rightarrow h = 2$.Then $k = \frac{2^{2}}{4} = 1$.The equation of the normal at $(2, 1)$ is:$y - 1 = -\frac{2}{2}(x - 2)$$y - 1 = -(x - 2)$$y - 1 = -x + 2$$x + y = 3$.The correct answer is $D$.

The normal to the curve $x^{2}=4y$ passing through $(1,2)$ is

Similar Questions

Where is the National Botanical Research Institute $(NBRI)$ located?

$ICBN$ stands for

The correct stability order for the following carbocations is:

$A$ steel wire can withstand a load up to $2940 ~N$. $A$ load of $150 ~kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module