Question

If the straight line x cosα + y sinα = p touches the curve `x^2/"a"^2 + y^2/"b"^2` = 1, then prove that a² cos²α + b² sin²α = p².

Answer 1

The given curve is `x^2/"a"^2 + y^2/"b"^2` = 1   ....(i)

And the straight line x cos a + y sin a = p

Differentiating equation (i) w.r.t. x, we get

`1/"a"^2 * 2x + 1/"b"^2 * 2y * "dy"/"dx"` = 0

⇒ `x/"a"^2 + y/"b"^2 "dy"/"dx"` = 0

⇒ `"dy"/"dx" = - "b"^2/"a"^2 * x/y`

So the slope of the curve = `(-"b"^2)/"a"^2 * x/y`

Now differentiating eq. (ii) w.r.t. x, we have

`cos alpha + sin alpha * "dy"/"dx"` = 0

∴ `"dy"/"dx" = (- cos alpha)/sinalpha`

= `- cot alpha`

So, the slope of the straight line = `- cot alpha`

If the line is the tangent to the curve, then

`(-"b"^2)/"a"^2 * x/y = - cot alpha`

⇒ `x/y = "a"^2/"b"^2 * cot alpha`

⇒ x = `"a"^2/"b"^2 cot alpha * y`

Now from equation (ii) we have x cos a + y sin a = p

⇒ `"a"^2/"b"^2 * cot alpha * y * cos alpha + y sin alpha` = p

⇒ `"a"^2 cot alpha * cos alpha y + "b"^2 sin alpha y = "b"^2"p"`

⇒ `"a"^2 cosalpha/sinalpha * cos alpha y + "b"^2 sin alpha y = "b"^2"p"`

⇒ `"a"^2 cos^2 alpha y + "b"^2 sin^2 alpha y = "b"^2 sin alpha "p"`

⇒ `"a"^2 cos^2 alpha + "b"^2 sin^2 alpha = "b"^2/y * sin alpha * "p"`

⇒ `"a"^2cos^2alpha + "b"^2 sin^2alpha = "p" * "p"`  ....`[because "b"^2/y sin alpha = "p"]`

Hence, a² cos²α + b² sin²α = p² 

Alternate method:

We know that y = mx + c will touch the ellipse

`x^2/"a"^2 + y^2/"b"^2` = 1 if c² = a²m² + b²

Here equation of straight line is x cos α + y sin α = p and that of ellipse is `x^2/"a"^2 + y^2/"b"^2` = 1

x cos α + y sin α = p

⇒ y sin α= – x cos α + p

⇒ y = `- x cosalpha/sinalpha + "P"/sinalpha`

⇒ y = `- x cot alpha + "P"/sinalpha`

Comparing with y = mx + c, we get

m = `- cot alpha` and c = `"P"/sinalpha`

So, according to the condition, we get c² = a²m² + b²

`"P"^2/(sin^2alpha) = "a"^2(- cot alpha)^2 + "b"^2`

 ⇒ `"P"^2/(sin^2alpha) = ("a"^2 cos^2alpha)/(sin^2alpha) + "b"^2`

⇒ p² = a² cos²α + b² sin²α

Hence, a² cos²α + b² sin²α = p² 

Hence proved.