Question
Question: The equation of the normal to the parabola \[{{x}^{2}}=8y\] whose slope is \[\dfrac{1}{m}\] A.\[y=...
The equation of the normal to the parabola x2=8y whose slope is m1
A.y=mx−2m−m3
B.x=mx−4m−2m3
C.x=my−4m−2m3
D.4y=mx+4m−2m3
Solution
Hint: Differentiate the equation of the parabola x2=8y with respect to x, we will get the slope of the tangent. We know that the product of slope of two perpendicular lines is -1. Since tangent and normal are perpendicular to each other. So, Slopeoftangent !!×!! Slope of Normal = -1 . Using this, we get the slope of normal and make it equal to m1 . Now, get the value of x. Put the value of x and in the equation x2=8y and get the value of y. Using the formula, (y−y1)=slope(x−x1) get the equation of the normal.
Complete step-by-step answer:
According to the question, it is given that,
Slope of normal = m1 ………………………(1)
First of all, we have to get the slope of the tangent. For that, we have to differentiate the equation of the parabola.
Differentiating the equation of the parabola x2=8y with respect to x, we get
dxd(x2)=dxd(8y)………………….(2)
We know the formula, dxdxn=nxn−1 .
Now, using this formula in equation (2), we get