0

Guest

Courses New

Let ax-y+7=0 be a chord of parabola x^2=28y meeting it at A and B ,and tangents at A and B meet at C.the locus of circumcentre of tr.ABC is

Let ax-y+7=0 be a chord of parabola x^2=28y
meeting it at A and B ,and tangents at A and B meet at C.the locus of circumcentre of tr.ABC is

Grade:12th pass

1 Answers

Latika Leekha
askIITians Faculty 165 Points
9 years ago
Hello student,
Given chord is ax – y + 7 = 0 and the parabola is x2 = 28y.
We first try to find out the points of intersection of the chord with the parabola as it would give us the end points of the chord.
So, substituting the value of y = ax+7 from the equation of the chord into the equation of the parabola we have,
x2 = 28(ax+7)
So, x = 14(a ± √a2+1)
y = x2/28
Let us denote the x and y cordinates as x*x** and y*y**.
Then the slope of the tangents at these two points would be given by
m*m** which would be equal to the value of dy/dx at x*x**.
Hence, m*m** = x*x**/14.
This clearly shows that the product of slopes is -1.
Hence, this means that the two tangents are perpendicular and therefore the triangle is a right angled triangle.
Circumcentre of a right angled traingle is given by the mid-point of the hypotenuse.
So, the coordinates of circumcentre are x$ = 14a
y$ = (x*2 + x**2)/ 28 = 7(2a2+1)

Think You Can Provide A Better Answer ?

Other Related Questions on Analytical Geometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More