Q70.The equation of the normal to the parabola, x2 = 8y at x = 4 is (1) x + 2y = 0 (2) x + y = 2 (3) x −2y = 0 (4) x + y = 6 y2
What This Question Tests
This question is a direct application of finding the equation of the normal to a parabola at a given point, requiring differentiation to find the tangent slope and then the normal slope.
Concepts Tested
Formulas Used
Equation of parabola x^2 = 4ay
dy/dx for tangent slope
Slope of normal = -1 / (slope of tangent)
Equation of line: y - y1 = m(x - x1)
📚 NCERT Sections This Tests
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9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
9.15 — Apply Mirror Equation And The Condition:
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9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
📋 Question Details
- Chapter
- Parabola
- Topic
- Equation of normal to a parabola
- Year
- 2012
- Shift
- 19 May Online
- Q Number
- Q70
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 11: Conic Sections
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