Skip to main content
Back to PARABOLA

JEE MATHEMATICS MCQ: PARABOLA

1 Easy
1 of 1
Analyzing +4 | -1
The number of real tangents that can be drawn to the curve y2+2xy+x2+2x+3y+1=0{y^2} + 2xy + {x^2} + 2x + 3y + 1 = 0 from the point (1,−2)\left( {1, - 2} \right) is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the point (1, -2) lies inside the parabola, and no real tangents can be drawn from an interior point.

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 →