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प्रश्न
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.
उत्तर
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.
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