Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Hence, the equation of the normal to the parabola y2 = 4x is y = mx – 2m – m3.
Now, if it passes through (h, k), then we have k = mh – 2m – m3
This can be written as m3 + m(2-h) + k = 0 …. (1)
Here, m1 + m2 + m3 = 0
m1m2 + m2m3 + m3m1 = 2 - h
Also, m1m2m3 = -k, where m1m2 = α
Now, this gives m3 = -k/α and this must satisfy equation (1).
Hence, we have (-k/α)3 + (-k/α)(2-h) + k = 0
Solving this, we get k2 = α2h – 2α2 + α3
And so we have y2 = α2x – 2α2 + α3
On comparing this equation with y2 = 4x we get
α2 = 4 and -2α2 + α3 = 0
This gives α = 2.
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !