Question

By taking a number of solutions, as in Arts. $39-41$, sketch the loci of the following equations :
1. $2x+3y=10$ 2. $4x-y=7$ 3. $x^2-2ax+y^2=0$
4. $x^2-4ax+y^2+3a^2=0$ 5. $y^2=x$ 6. $3x=y^2-9$
7. $\frac{x^2}{4}+\frac{y^2}{9}=1$

Answer 1

$1.$ $2x+3y=10$

$y=0$$\Rightarrow x=5$

$A(5,0)$

$x=0\Rightarrow y=\frac{10}{3}B(0,\frac{10}{3})$

Represents a straight line joining $AB$

$-----------------------------------------$

$2.$ $4x-y=7$

$y=0\Rightarrow x=\frac{7}{4}A(\frac{7}{4},0)x=0\Rightarrow y=-7B(0,-7)$

Represents a straight line joining $AB$

$-----------------------------------------$

$3.$ $x^2-2ax+y^2=0$

$x^2-2ax+a^2+y^2=a^2{(x-a)}^2+y^2=a^2$

represents a circle with centre $(a,0)$ and radius $a$

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$4.$ $x^2-4ax+y^2+3a^2=0$

$x^2-4ax+4a^2-{4a}^2+y^2+3a^2=0{(x-2a)}^2+y^2=a^2$

Represents a circle with centre $(2a,0)$ and radius $a$

$-----------------------------------------$

$5.$ $y^2=x$

Comparing with the general equation $y^2=4ax$

$a=\frac{1}{4}$

Represents a parabola with focus $(\frac{1}{4},0)$ and vertex $(0,0)$

$------------------------------------------$

$6.$ $3x=y^2-9$

$2x=y^2-9y^2=3x+9y^2=3(x+3)$

Comparing with the general equation ${(y-k)}^2=4a(x-h)$

$4a=3a=\frac{3}{4}$

Focus: $(\frac{3}{4}-3,0)=(-\frac{9}{4},0)$

Vertex: $(0-3,0)=(-3,0)$

So the equation represents a parabola.

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$7.$ $\frac{x^2}{4}+\frac{y^2}{9}=1$

$a=2,b=3$

$b>a$

Clearly it represents an ellipse with $y$ as major axis.