If the tangent drawn at P ( t = 4 π ) to the curve x = sec 2 t , y = cot t meets the curve again at R, then, PR = \_\_\_\_

At t = 4 π , x = 2 , y = 1 ⇒ P is ( 2 , 1 ) d x d y t = π /4 = 2 s e c 2 t t a n t − c s c 2 t = − 2 1 ∴ tangent at P ( 2 , 1 ) is, y = 2 4 − x Eliminating 't', curve equation is x = 1 + y 2 1 ; solving with the tangent, R ( 5 , − 2 1 ) ⇒ PR = 2 3 5

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediumsingle choice

If the tangent drawn at $P (t = \frac{π}{4})$ to the curve $x = sec^{2} t, y = cot t$ meets the curve again at R, then, PR = \_\_\_\_

Options:

Answer:

Solution:

At $t = \frac{π}{4}, x = 2, y = 1 \Rightarrow P$ is $(2, 1)$ $\frac{d y}{d x}_{t = π /4} = \frac{- c s c ^{2} t}{2 s e c ^{2} t t a n t} = - \frac{1}{2}$

∴ tangent at $P (2, 1)$ is, $y = \frac{4 - x}{2}$ Eliminating 't', curve equation is $x = 1 + \frac{1}{y ^{2}}$ ; solving with the tangent, $R (5, - \frac{1}{2}) \Rightarrow PR = \frac{3}{2} 5$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Tangents & Normals

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...