Q.If the curves $ax^2 + by^2 = 1$ and $a'x^2 + b'y^2 = 1$ intersect orthogonally, then prove that $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{1}{a'} - \dfrac{1}{b'}$.

The tangent at $x=3$ is $10x - y = 8$.

Given $y = x^2+4x+1$. The slope is

$\frac{dy}{dx} = 2x+4.$

At $x=3$: slope $= 2(3)+4 = 10$, and $y = 9+12+1 = 22$, so the point is $(3,22)$.

Equation of the tangent:

$y - 22 = 10(x-3) \Rightarrow y = 10x - 30 + 22 = 10x - 8,$

i.e. $10x - y = 8$.

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.

For $y^2=4ax$, the point $(at^2,2at)$ satisfies $x=at^2$, $y=2at$.

$\dfrac{dx}{dt}=2at$, $\dfrac{dy}{dt}=2a$

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2a}{2at}=\dfrac{1}{t}$

The slope of the tangent at a point equals $\dfrac{dy}{dx}$ evaluated at that point.

Given $y = x^3 - x$

$\dfrac{dy}{dx} = 3x^2 - 1$

At $x = 2$: $\dfrac{dy}{dx} = 3(2)^2 - 1 = 12 - 1 = 11$

Find $dy/dx$ for the parametric curve and evaluate at $\theta=\pi/4$.

$x = a\cos^3\theta \Rightarrow \dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$.

$y = a\sin^3\theta \Rightarrow \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$.

$\dfrac{dy}{dx} = \dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\dfrac{\sin\theta}{\cos\theta} = -\tan\theta$.

At $\theta = \pi/4$: $-\tan(\pi/4) = -1$.

Matching the curve's slope to the tangent's slope $1$ gives the point $(2,-9)$.

$y=x^3-11x+5\Rightarrow \dfrac{dy}{dx}=3x^2-11$. The tangent line $y=x-11$ has slope $1$, so $3x^2-11=1\Rightarrow x^2=4\Rightarrow x=\pm2$.

At $x=2$: $y=8-22+5=-9$, giving $(2,-9)$; check it lies on the tangent: $2-11=-9$ ✓.

At $x=-2$: $y=-8+22+5=19$, but the tangent gives $-2-11=-13\ne19$, so this point is rejected.