A curve $C$ is defined by the parametric equations $x={{t}^{2}}$, $y={{t}^{3}}$. Prove that th

Question

A curve $C$ is defined by the parametric equations $x={{t}^{2}}$, $y={{t}^{3}}$. Prove that the equation of the tangent at the point $\left( {{t}^{2}},{{t}^{3}} \right)$ on the curve is $2y-3tx+{{t}^{3}}=0$.[3] This tangent passes through a fixed point $\left( a,b \right)$. Explain why there cannot be more than $3$ tangents through $\left( a,b \right)$.[1] The tangent at the point $P$ when $t=2$ meets the curve again at the point $Q$ where $t=k$. Find the value of $k$.[3] Find the equation of the normal at the point $P$.[2] Sketch the curve $C$, indicating the intersection point(s) with the axes.[2] By sketching the tangent and normal at the point $P$ on the sketch in (v), explain why ${{\tan }^{-1}}\left( 3 \right)+{{\tan }^{-1}}\left( \frac{1}{3} \right)=\frac{\pi }{2}$.[2]

Accepted Answer

(ii) A cubic equation has at most $3$ real roots. Given that $\left( a,b \right)$ is a fixed point, the equation $2b-3at+{{t}^{3}}=0$ is a cubic equation in terms of $t$. Hence there are at most $3$ real values of $t$ for a fixed value of $x$ and $y$ and therefore at most $3$ tangents can pass through the fixed point $\left( a,b \right)$. (iii) $k=-1$ . The tangent will meet the curve again at $k=-1$. (iv) $y=-\frac{1}{3}x+\frac{28}{3}$ (vi) From the graph, $\alpha +\beta =\frac{\pi }{2}$, $\alpha ={{\tan }^{-1}}\left( 3 \right)$, $\beta ={{\tan }^{-1}}\left( \frac{1}{3} \right)$, since $\beta $ is acute $\therefore {{\tan }^{-1}}\left( 3 \right)+{{\tan }^{-1}}\left( \frac{1}{3} \right)=\frac{\pi }{2}$.

2023 SAJC Promo Q9

2023 SAJC Promo Q9

Video Solution:

Solution:

Solution locked