Question

8. If b1 and b2 be the length of perpendicular drawn from the points
$\sqrt{ {a }^{2} - {b}^{2} }$
and
$- \sqrt{ {a}^{2} - {b}^{2} }$
upon the line
${x \cos( \beta ) \div a} + y \sin( \beta \div b) = 1$
Then show that b1×b2=
${b}^{2}$

​

Answer 1

Appropriate Question :-

If $b_1\:and\:b_2$ be the length of perpendicular drawn from the points $(\sqrt{ {a }^{2} - {b}^{2} },0)$ and $(- \sqrt{ {a}^{2} - {b}^{2} }, 0)$ upon the line

${x \cos( \beta ) \div a} + y \sin( \beta) \div b = 1$

Then show that $b_1 \times b_2= {b}^{2}$

$\large\underline{\sf{Solution-}}$

Given equation of line is

$\rm \: \dfrac{x}{a}cos\beta + \dfrac{y}{b}sin\beta = 1$

can be rewritten as

$\rm \: bxcos\beta + aysin\beta = ab$

$\rm \: bxcos\beta + aysin\beta - ab = 0 - - - (1)$

Now,

$\rm \: b_1$

$\rm \: = distance \: from \: ( \sqrt{ {a}^{2} - {b}^{2}}, \: 0) \: on \: line \: (1)$

$\rm \: = \: \bigg |\dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta + 0 - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

$\rm \: = \: \bigg |\dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

Now, Consider

$\rm \: b_2$

$\rm \: = distance \: from \: ( - \sqrt{ {a}^{2} - {b}^{2}}, \: 0) \: on \: line \: (1)$

$\rm \: = \: \bigg |\dfrac{ - b \sqrt{ {a}^{2} - {b}^{2}} cos\beta + 0 - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

$\rm \: = \: \bigg |\dfrac{ - b \sqrt{ {a}^{2} - {b}^{2}} cos\beta - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

$\rm \: = \: \bigg |\dfrac{ - (b \sqrt{ {a}^{2} + {b}^{2}} cos\beta + ab)}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

$\rm \: = \: \bigg |\dfrac{b \sqrt{ {a}^{2} + {b}^{2}} cos\beta + ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg |$

Now, Consider

$\rm \: b_1 \times b_2$

$\rm \: = \bigg |\dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg| \times \bigg |\dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta + ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg|$

$\rm \: = \bigg |\dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta - ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } } \times \dfrac{b \sqrt{ {a}^{2} - {b}^{2}} cos\beta + ab}{ \sqrt{ {b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } }\bigg|$

$\rm \: = \bigg |\dfrac{ {b}^{2}({a}^{2} - {b}^{2}) cos^{2} \beta - {a}^{2} {b}^{2} }{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \bigg |\dfrac{ {b}^{2}[({a}^{2} - {b}^{2}) cos^{2} \beta - {a}^{2}]}{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \bigg |\dfrac{ {b}^{2}[{a}^{2} {cos}^{2}\beta-{b}^{2}cos^{2} \beta - {a}^{2}]}{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \bigg |\dfrac{ {b}^{2}[ - {a}^{2}(1 - {cos}^{2}\beta)-{b}^{2}cos^{2} \beta ]}{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \bigg |\dfrac{ {b}^{2}[ - {a}^{2}{sin}^{2}\beta-{b}^{2}cos^{2} \beta ]}{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \bigg |\dfrac{ - {b}^{2}[{a}^{2}{sin}^{2}\beta + {b}^{2}cos^{2} \beta ]}{{b}^{2} {cos}^{2} \beta + {a}^{2} {sin}^{2}\beta } \bigg|$

$\rm \: = \: | - {b}^{2} |$

$\rm \: = \: {b}^{2}$

Hence,

$\rm\implies \:\rm \: \boxed{ \rm{ \: \:b_1 \times b_2 = {b}^{2} \: \: }}$

$\rule{190pt}{2pt}$

Formula Used :-

The perpendicular distance (d) drawn from the point (p, q) on the line ax + by + c = 0 is given by

$\boxed{ \rm{ \:d \: = \: \bigg | \frac{ap + bq + c}{ \sqrt{ {a}^{2} + {b}^{2} } }\bigg | \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.