# Straight Line equation problem

Show that the equation of the line passing through $(a\cos^3\theta,a\sin^3\theta)$ and perpendicular to the line $x\sec\theta+y\csc \theta=a$ is $x\cos\theta-y\sin\theta= a\cos2\theta$

My attempt:

I converted the second line to the intercept form:

$\dfrac{x}{a\cos\theta}+\dfrac{y}{a\sin\theta}=1$

Thus slope of this line is: $\dfrac{y\cos\theta}{x\sin\theta}$

And,

slope of it's perpendicular is $\dfrac{-x\sin\theta}{y\cos\theta}$

Using point slope form, ($y-y_1=mx-x_1$)

Equation of required line is:

$y^2\cos\theta - a\sin^3\theta y\cos\theta = -x^2 \sin\theta +ax\sin\theta cos^3\theta$

I am unable to continue from here.How do I reach the desired solution from here?

PS: I know $\cos2\theta$'s formula too.

• What happens when you plug in for $x,y$ the values $a \cos^3 \theta, a\sin^3 \theta$ (the point the line is to pass through)? – coffeemath Aug 7 '17 at 9:44
• Oh. I think I have made a mistake in finding the slope. – user342531 Aug 7 '17 at 9:49
• It's better to follow your solution from line's slope $\dfrac{a\sin\theta}{a\cos\theta}$. – Nosrati Aug 7 '17 at 10:07
• Yes, I had made a mistake in finding the slope. I used Dr's method for the same and reached the desired solution – user342531 Aug 7 '17 at 10:41
• Your derivation doesn’t work for the case that $\theta$ is a multiple of $\pi/4$: you’re dividing by zero right off the bat. – amd Aug 7 '17 at 19:08

we have $$P(a\cos^3(\theta);a\sin^3(\theta))$$ and $$y=mx+n$$ the straight line, then we plug in the coordsinate of P: $$y=m(x-a\cos^3(\theta))+a\sin^3(\theta)$$ (I) the given equation has the form $$y=-\frac{\sec(\theta)}{\csc(\theta)}x+\frac{a}{\csc(\theta)}$$ then is $$m=\frac{\csc(\theta)}{sec(\theta)}$$ the searched slope then you must plug in $m$ in (I)
• when $$m$$ is the slope of a line then $$\frac{-1}{m}$$ the slope of the perpendicular line – Dr. Sonnhard Graubner Aug 7 '17 at 10:01