Question

If P1 and P2 be the lengths of perpendiculars from the origin on the tangent and normal to the curve x2/3  y2/3  a2/ 3 respectively, the value of 2 2 4P1  P2 is

Answer 1

4 p1² + p2² = a²

Step-by-step explanation:

Please refer to the attached picture for the question and the graph for reference.

p1² + p2² = OA² (Pythagoras Theorem)

Tangent equation: x tansθ + y - (aCos^3θ tanθ + asin^3θ ) = 0

Distance p1 from origin, we get: p1 = a/2. Sinθ

p2 = a.Cosθ

4 p1² + p2² = 4. a²/4.Sin²θ + a².Cos²θ

    = a².Sin²θ + a².Cos²θ

    = a² (Sin²θ + Cos²θ)

4 p1² + p2² = a²

Answer 2

The value of

4P1² + P2² = a²

Step-by-step explanation:

The given curve is

$x^{2/3}+y^{2/3}=a^{2/3}$

In parametric form, any point on the curve will be

$(a\cos^3\theta, a\sin^3\theta)$

The tangent at any point of the curve is

$\frac{dy}{dx}$

Differentiation the given curve equation w.r.t. x

$\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0$

$[\implies \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}$

$[\implies \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}$

slope at point $(a\cos^3\theta, a\sin^3\theta)$

$\frac{dy}{dx}\Bigr|_{(a\cos^3\theta, a\sin^3\theta)}=-\frac{(a\sin^3\theta)^{1/3}}{(a\cos^3\theta)^{1/3}}$

$\implies \frac{dy}{dx}\Bigr|_{(a\cos^3\theta, a\sin^3\theta)}=-\frac{\sin\theta}{\cos\theta}$

$\implies \frac{dy}{dx}\Bigr|_{(a\cos^3\theta, a\sin^3\theta)}=-\tan\theta$

Therefore, the slope of normal at the given point will be

$-\frac{1}{(-\tan\theta)}$         (We know that $m_1m_2=-1$)

$=\cot\theta$

Now

Equation of the tangent at point $(a\cos^3\theta, a\sin^3\theta)$

$(y-a\sin^3\theta)=-\tan\theta (x-a\cos^3\theta)$

or, $(y-a\sin^3\theta)=-\frac{\sin\theta}{\cos\theta} (x-a\cos^3\theta)$

or, $y\cos\theta-a\sin^3\theta\cos\theta=-\sin\theta x+a\cos^3\theta\sin\theta$

or, $y\cos\theta+\sin\theta x-a(\cos^3\theta\sin\theta+\sin^3\theta\cos\theta)=0$

The length of the perpendicular on this line

$P_1=\frac{a(\cos^3\theta\sin\theta+\sin^3\theta\cos\theta)}{\sqrt{\cos^2\theta+\sin^2\theta}}$

$\implies P_1=a\cos\theta\sin\theta(\cos^2\theta+\sin^2\theta)$

$\implies P_1=a\cos\theta\sin\theta$

Similarly, equation of normal at the point

$(y-a\sin^3\theta)=\cot\theta (x-a\cos^3\theta)$

or, $(y-a\sin^3\theta)=\frac{\cos\theta}{\sin\theta} (x-a\cos^3\theta)$

or, $y\sin\theta-a\sin^4\theta=\cos\theta x-a\cos^4\theta$

or, $\cos\theta x-\sin\theta y-a(\cos^4\theta-\sin^4\theta)=0$

or, $\cos\theta x-\sin\theta y-a[(\cos^2\theta)^2-(\sin^2\theta)^2]=0$

or, $\cos\theta x-\sin\theta y-a[(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)]=0$

or, $\cos\theta x-\sin\theta y-a\cos2\theta=0$

The length of the perpendicular on this normal line

$P_2=\frac{|a\cos2\theta|}{\sqrt{\cos^2\theta+\sin^2\theta}}$

$\implies P_2=a\cos2\theta$

Now

$4P_1^2+P_2^2$

$=4a^2(\cos\theta\sin\theta)^2+a^2\cos^22\theta$

$=a^2(2\cos\theta\sin\theta)^2+a^2\cos^22\theta$

$=a^2(\sin^22\theta+\cos^22\theta)$

$=a^2$

Hope this answer is helpful.

Know More:

Q: If p1 and P2 be the lengths of the perpendiculars from origin on xsec@ + ycosec@ = a and xcos@ - ysin@ = acos2@ , find 4p1 Square + p2 square

Click Here: https://brainly.in/question/1326097

Q: f p and q are the lengths of perpendiculars from the origin to the lines x cos ? – y sin ? = k cos 2? and x sec ?+ y cosec ? = k, respectively, prove that p^2 + 4q^2 = k^2

Click Here: https://brainly.in/question/1763861