1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The normal chord of a parabola y2=4ax at the point whose ordinate is equal to the abscissa then angle subtended by normal chord at the focus is:

A
π4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
tan12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
tan12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
π2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D π2Point whose ordinate equal to abscissa will be such that x=y,but (x,y) lies on y2=4ax⇒ x2=4ax⇒ x=0 or x=4aSince at (0,0) we cannot draw a normal chord, hence point at which normal chord is drawn is P(4a,4a).Focus of the given parabola y2=4ax is S(a,0).Equation of the normal chord at (4a,4a) will bey−4ax−4a=−2⇒ 2x+y=12a is the equation of normal.To find the points of intersection of normal with the parabola , we need tosolve the equation (12a−2x)2=4ax⇒ (6a−x)2=ax⇒ 36a2−12ax+x2=ax⇒ x2−13ax+36a2=0⇒ (x−4a)(x−9a)=0⇒ x=4a or 9aP(4a,4a) and Q(9a,−6a) are the end points of the normal chord.Slope of line PS=(4a−0)(4a−a)=43Slope of line QS=(−6a−0)(9a−a)=−34We can observe that product of (SlopeofPS)×(SlopeofQS)=−1⇒ ∠PSQ=90o.∴ Normal chord at the point whose ordinate is equal to abscissa subtends an angle of π2 at the focus.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Parabola
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program