Skip to main content
Back to PARABOLA

JEE MATHEMATICS MCQ: PARABOLA

1 Easy
1 of 1
Application +4 | -1
Tangent to the parabola y=x2+6y=x^2+6 at (1,7)(1,7) touches the circle x2+y2+16x+12y+c=0x^2+y^2+16 x+12 y+c=0 at the point
View answer and explanation
Correct option: C
(−6,−7)(-6,-7)
Option C ((-6,-7)) is correct because it is the point of tangency between the line and the circle.

More Questions from PARABOLA

Practice All Chapter MCQs →
The equation of a common tangent to the parabolas y=x2y=x^2 and y=−(x−2)2y=-(x-2)^2 is A point PP moves such that slope of one tangent drawn from it to parabola y2=16xy^2=16 x is four times the slope of other tangent, then the length of latus rectum of the conic represented by locus of PP, is Let P,Q,RP,Q,R be three points on parabola y2=4x{y^2} = 4x and normal at PP and RR meet at QQ, then the locus of the mid-point of the chord PRPR is a parabola whose vertex is at The tangents at the end points of any chord through (1,0)(1,0) to the parabola y2+4x=8y^2+4 x=8 intersect The focus of the parabola x2=−16yx^2=-16 y is If II denotes the semi-latusrectum of the parabola y2=4ax{y^2} = 4ax and SPSP and SQSQ denote the segments of any focal chord PQ,SPQ,S being the focus, then SP,lSP,l and SQSQ are in the relation
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →