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.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !