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Solution :

Let foot of the perpendicular from vertex on tangent be M(h,k). <br> Slope of OM `=(k)/(h)` <br> `:.` Slope of tangent `=-(h)/(k)` <br> Thus, equation of tangent is <br> `y-k=-(h)/(k)(x-h)` <br> `ory=-(h)/(k)x+(h^(2)+k^(2))/(k)` <br> Comparing with `y=mx+(a)/(m)`, we have <br> `m=-(h)/(k)and(a)/(m)=(h^(2)+k^(2))/(k)` <br> `:." "-(ak)/(h)=(h^(2)+k^(2))/(k)` (eliminating m) <br> So, `x(x^(2)+y^(2))+ay^(2)=0` is the required locus.**meaning of function**

**Modulus; signum; greatest integer and fractional part function**

**representation of function**

**Physical interpretation**

**Domain codomain and range of function**

**Graph of function: Vertical line test**

**Identity function**

**Explain Constant function with graph**

**Modulus function and properties**

**Greatest integer function and properties**