If the tangents and normals at the extremities of a focal chord of any standard parabola intersect at S≡(x1,y1) and R≡(x2,y2) respectively, then
A
If the equation of parabola is y2=±4ax, then y1=y2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
If the equation of parabola is x2=±4ay, then x1=x2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
the circle with focal chord as diameter will always pass through the points R and S
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
the circle with RS as diameter will always pass through extemities of focal chord
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D the circle with RS as diameter will always pass through extemities of focal chord Let P(at21,2at1) and Q(at22,2at2) be the extremities of a focal chord of the parabola y2=4ax. The tangents at P and Q intersect at S≡(at1t2,a(t1+t2))⇒S≡(−a,a(t1+t2)). [∵PQ is a focal chord] ∴x1=−a and y1=a(t1+t2)
The normals at P and Q intersect at R≡(2a+a(t21+t22+t1t2),−at1t2(t1+t2))⇒R≡(a+a(t21+t22),a(t1+t2)) ∴x2=a+a(t21+t22) and y2=a(t1+t2)
Clearly, y1=y2
Similarly it can be proved for the parabola y2=−4ax
and in same way if we consider the parabola of the form x2=±4ay, then we get x1=x2
Now from above figure in the quadrilateral PRQS ∠P+∠Q=∠S+∠R=180 ∴PRQS will be cyclic quadrilateral