Let C be a curve given by y(x) = 1 + $$\sqrt {4x - 3} ,x > {3 \over 4}.$$ If P is a point on C, such that the tangent at P has slope $${2 \over 3}$$, then a point through which the normal at P passes, is :
JEE · Math · previous-year question
- A.(2, 3)
- B.(4, $$-$$3)
- C.(1, 7)correct
- D.(3, $$-$$ 4),
Answer
C. (1, 7)
Explanation
Given, y = 1 + $$\sqrt {4x - 3} $$ $$ \therefore $$ $${{dy} \over {dx}}$$ = $${1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}$$ $$ \Rightarrow $$ 4x $$-$$ 3 = 9 $$ \Rightarrow $$ x = 3 $$ \therefore $$ y = 1 + $$\sqrt {12 - 3} $$ = 4 $$ \therefore $$ Equation of normal at point P(3,4) y $$-$$ 4 = $$-$$ $${3 \over 2}$$ (x $$-$$ 3) $$ \Rightarrow $$ 2y $$-$$ 8 = $$-$$ 3x + 9 $$ \Rightarrow $$ 3x + 2y $$-$$ 17 = 0
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