Question

If a straight line lx+my+n= 0 touches the hyperbola x²/a²- y²/b² = 1 ,show that a²l²- b²m²= n²​

Answer 1

EXPLANATION.

→ straight line = lx + my + n = 0

→ touches the hyperbola = x²/a² - y²/b² = 1

→ To show that = a²l² - b²m² = n.

→ equation of line → y = mx + c

→ lx + my + n = 0

→ my = - lx - n

→ y = -lx/m - n/m

→ conditions of tangent for the line y = mx + c

to be tangent to hyperbola x²/a² - y²/b² = 1.

→ c² = a²m² - b²

→ put the equation in this conditions.

→ ( -n/m)² = a²( -l/m)² - b²

→ n²/m² = a²l²/m² - b²

→ n² = a²l² - m²b² = Hence proved.

More information.

→ equation of tangent.

(1) → slope form.

→ c² = a²m² - b²

→ c = ± √a²m² - b²

→ equation of tangent → y = mx ± √ a²m² - b².

(2) → point form.

→ x²/a² - y²/b² = 1

→ xx¹/a² - yy¹/b² = 1 = point form.

(3) → parametric form.

→ x²/a² - y²/b² = 1

→ xx¹/a² - yy¹/b² = 1

→ let the parametric coordinate be :-

( a sec ø, b tan ø)

→ put the parametric point in equation.

→ x. a sec ø / a² - y. b tan ø / b² = 1.

$\sf : \implies \: \dfrac{x \sec( \theta) }{a} - \dfrac{y \tan( \theta) }{b} = 1$