Question

If the chord y=mx+c subtends a right angle at the vertex of the parabola

Answer 1

The question is incomplete we need to find out the value of c.

We will imagine parabola to be x² = 4ay and for this the end point will be (2at1, at1²)

(2at2,at2²)

The lines joining these point of the vertex will give this product: (0,0) is -1

(at1²−0)/(2at1−0)∗(at2²−0)/(2at2−0)=−1

this equation will give t1 x t2 = -4

For this the slop of the chord will be  (t1+t2)/2

and for this following will be the equation:

Y−at12=(t1+t2)/2∗(x−2at1)

we will solve this:

Y=(t1+t2)∗x/2−a∗t1∗t2

so for this the value of c will be 4a

so for this if the chord passes through the focus (0,a) then t1 x t2 = -1

the result for this will be 90° at the vertex.